MA162: Finite mathematics . Jack Schmidt University of Kentucky - PowerPoint PPT Presentation

. MA162: Finite mathematics . Jack Schmidt University of Kentucky February 25, 2013 Schedule: HW 3.1-3.3, 4.1 (Late) HW 2.5-2.6 due Friday, Mar 01, 2013 Exam 2, Monday, Mar 04, 2013, from 5pm to 7pm HW 5.1 due Friday, Mar 08, 2013 Spring Break, Mar 09-17, 2013 HW 5.2-5.3 due Friday, Mar 22, 2013 Today we will cover 2.6: matrix inverse, and use it to make a spreadsheet solve LPPs

2.5: An easy multiplication, the identity There is a matrix that doesn’t change things when it multiplies against them:  1 0 0 0   11 12 13  . . . 0 1 0 0 21 22 23      · . . .     0 0 1 0 31 32 33    . . . 0 0 0 1 41 42 43 . . .   1 · 11 + 0 · 21 + 0 · 31 + 0 · 41 12 13 . . . 0 · 11 + 1 · 21 + 0 · 31 + 0 · 41 22 23   = . . .   0 · 11 + 0 · 21 + 1 · 31 + 0 · 41 32 33   . . . 0 · 11 + 0 · 21 + 0 · 31 + 1 · 41 42 43 . . . Make sure the size of the matices match though!

2.5: An easy multiplication, the identity There is a matrix that doesn’t change things when it multiplies against them:  1 0 0 0   11 12 13  . . . 0 1 0 0 21 22 23      · . . .     0 0 1 0 31 32 33    . . . 0 0 0 1 41 42 43 . . .   1 · 11 + 0 · 21 + 0 · 31 + 0 · 41 12 13 . . . 0 · 11 + 1 · 21 + 0 · 31 + 0 · 41 22 23   = . . .   0 · 11 + 0 · 21 + 1 · 31 + 0 · 41 32 33   . . . 0 · 11 + 0 · 21 + 0 · 31 + 1 · 41 42 43 . . . Make sure the size of the matices match though!

2.5: An easy multiplication, the identity There is a matrix that doesn’t change things when it multiplies against them:  1 0 0 0   11 12 13  . . . 0 1 0 0 21 22 23      · . . .     0 0 1 0 31 32 33    . . . 0 0 0 1 41 42 43 . . .   1 · 11 + 0 · 21 + 0 · 31 + 0 · 41 12 13 . . . 0 · 11 + 1 · 21 + 0 · 31 + 0 · 41 22 23   = . . .   0 · 11 + 0 · 21 + 1 · 31 + 0 · 41 32 33   . . . 0 · 11 + 0 · 21 + 0 · 31 + 1 · 41 42 43 . . . Make sure the size of the matices match though!

2.5: An easy multiplication, the identity There is a matrix that doesn’t change things when it multiplies against them:  1 0 0 0   11 12 13  . . . 0 1 0 0 21 22 23      · . . .     0 0 1 0 31 32 33    . . . 0 0 0 1 41 42 43 . . .   11 12 13 . . . 21 22 23   = . . .   31 32 33   . . . 41 42 43 . . . Make sure the size of the matices match though!

2.5: Multiplication to solve one equation 2 3 · 3 2 = 1 To solve 3 2 x = 9 we just multiply by 2 3 x = 2 3 9 = 6 We are multiplying both sides by 2 3 Left side turns out nice and boring: 2 3 · ( 3 2 x ) = ( 2 3 · 3 2 ) x = (1) x = x

2.5: Multiplication to solve a system Matrix version: ( 1 ( 1 2 ) ( 3 − 2 ) 0 ) · = 1 3 − 1 1 0 1 To solve: ( 5 ( x ( ) ) ) 3 − 2 · = − 1 1 7 y Just multiply: ( x ( 1 ( 5 ( (1)(5) + (2)(7) ( 19 ) ) ) ) ) 2 = · = = y 1 3 7 (1)(5) + (3)(7) 26

2.6: Matrix division There are several ways to do matrix division, see book for tricks We’ll cover one systematic, basically easy way And we already know it , we just use RREF: If you know A and B , then to solve AX = B put the augmented matrix ( A | B ) into RREF as ( I | X ) In other words, RREF ( A | B ) = ( I | X ) inverses are solving AX = I , X = A − 1 , so we use RREF there too

2.6: Using RREF to solve the system ( 5 ( x ( ) ) ) 3 − 2 · = − 1 1 7 y Make augmented matrix and RREF ( − 1 ( 3 − 2 5 ) 1 7 ) R 1 ↔ R 2 R 2 +3 R 1 − − − − → − − − − − → − 1 1 7 3 − 2 5 ( − 1 ( − 1 ( 1 1 7 ) 0 − 19 ) 0 19 ) R 1 − R 2 − R 1 − − − − → − − → 0 1 26 0 1 26 0 1 26 Find inverse is almost exactly the same ( − 1 ( ) ) 3 − 2 1 0 1 0 1 R 1 ↔ R 2 R 2 +3 R 1 − − − − → − − − − − → − 1 1 0 1 3 − 2 1 0 ( − 1 ( − 1 ( 1 1 0 1 ) 0 − 1 − 2 ) 0 1 2 ) R 1 − R 2 − R 1 − − − − → − − → 0 1 1 3 0 1 1 3 0 1 1 3

2.6: Why is the inverse useful? The inverse allows you to solve AX = B using matrix multiplication instead of RREF A − 1 A = I A − 1 AX = IX = X If AX = B , then multiply both sides on the left by A − 1 then A − 1 AX = A − 1 B so X = A − 1 B Multiply by the inverse does the same thing as the long RREF Of course to find the inverse, we use RREF

4.1: Using the inverse to calculate RREF If a square matrix has an inverse, its RREF is the identity, very dull To find the RREF of a non-square matrix, we multiply by the inverse of a submatrix If the matrix is wide, then we choose one column per row to be in the submatrix We multiply the original by the inverse of the submatrix Those columns form the submatrix become pivots (one 1 and the rest 0s) The others change so that the X,Y,Z solutions don’t change

4.1: Using the inverse to calculate RREF A simple way to do the simplex algorithm in google docs The submatrix is specified by putting 0 or 1 below the column (0 is FREE, 1 is PIVOT) The formula for the submatrix is FILTER( Matrix, PivotList ) The formula for the RREF is: MMULT(MINVERSE(FILTER(Matrix,PivotList)),Matrix) See this google doc for an example from the practice exam. It also calculates the ratios to help find pivot rows